$\begin{aligned} &f(a)=\sqrt{a} \\\\ &g(b)=\dfrac{1}{b} \end{aligned}$ $(g\circ f) (0.25)=$
Let's start by rewriting $(g\circ f) (0.25)$ as $g(f(0.25))$. When evaluating composite functions, we work our way inside out. To evaluate $g(f(0.25))$, let's first evaluate $f(0.25)$. Then we'll plug that result into $g$ to find our answer. Let's evaluate $f({0.25})$. $\begin{aligned}f(a)&=\sqrt{a}\\\\ f({0.25})&=\sqrt{{0.25}}~~~~~~~~~~\text{Plug in }a={0.25}\\\\ &=\sqrt{\dfrac{1}{4}}\\\\ &={\dfrac{1}{2}}\end{aligned}$ We now know that $g(f({0.25}))$ is the same as $g\left({\dfrac{1}{2}}\right)$ because $f({0.25}) = {\dfrac{1}{2}}$. Let's evaluate $g\left({\dfrac{1}{2}}\right)$. $\begin{aligned}g(b)&=\dfrac{1}{b}\\\\ g\left({{\dfrac{1}{2}}}\right)&=\dfrac{1}{{\dfrac{1}{2}}}~~~~~~~~~~\text{Plug in }b={\dfrac{1}{2}}\\\\\\ &=2\\\\\\\\\end{aligned}$ The answer: $(g\circ f)(0.25) =2$